Print all subsets with given sum
Last Updated :
17 Mar, 2025
Given an array arr[] of non-negative integers and an integer target. The task is to print all subsets of the array whose sum is equal to the given target.
Note: If no subset has a sum equal to target, print -1.
Examples:
Input: arr[] = [5, 2, 3, 10, 6, 8], target = 10
Output: [ [5, 2, 3], [2, 8], [10] ]
Explanation: We need to find all subsets of arr[] that sum up to target = 10.
Subset [5, 2, 3] – Sum = 5 + 2 + 3 = 10
Subset [2, 8] – Sum = 2 + 8 = 10
Subset [10] – Sum = 10
Input: arr[] = [5, 7, 8], target = 3
Output: [ [-1] ]
Explanation: There are no subsets of the array that sum up to the target 3.
Input: arr[] = [35, 2, 8, 22], target = 0
Output: [ [ ] ]
Explanation: The empty subset is the only subset with a sum of 0.
Using Recursion – O(2^n) Time and O(n) Space
The idea is to use recursion to explore all possible subsets of the given array. We either include or exclude each element while keeping track of the remaining target sum. If we reach the end of the array and the target becomes 0, we store the valid subset. Otherwise, we backtrack and explore other possibilities.
C++
// C++ Code to find subsets with sum equal
// to target using recursion
#include <bits/stdc++.h>
using namespace std;
void findSubsets(vector<int>& arr, int index,
int target,vector<int>& curr,
vector<vector<int>>& result) {
if (index >= arr.size()) {
// If we reach the end and the target
// becomes 0, we found a valid subset
if (target == 0) {
result.push_back(curr);
return;
}
// Otherwise, we return as no valid
// subset is found
return;
}
// Include current element in subset
curr.push_back(arr[index]);
findSubsets(arr, index + 1,
target - arr[index], curr, result);
// Backtrack and exclude the current element
curr.pop_back();
findSubsets(arr,
index + 1, target, curr, result);
}
// Function to find all subsets summing to target
vector<vector<int>> perfectSum(vector<int>& arr,
int target) {
vector<vector<int>> result;
vector<int> curr;
findSubsets(arr, 0, target, curr, result);
return result;
}
// Function to print subsets in required format
void print2dArray(vector<vector<int>>& arr) {
if (arr.empty()) {
// No valid subsets found
cout << "-1\n";
return;
}
int row, col;
for (row = 0; row < arr.size(); row++) {
cout << "[";
for (col = 0; col < arr[row].size(); col++) {
cout << arr[row][col];
if (col != arr[row].size() - 1) {
cout << ", ";
}
}
cout << "]";
if(row < arr.size() - 1) cout << ", ";
}
}
int main() {
vector<int> arr = {5, 2, 3, 10, 6, 8};
int target = 10;
// Find subsets and print result
vector<vector<int>> result
= perfectSum(arr, target);
print2dArray(result);
return 0;
}
// Java Code to find subsets with sum equal
// to target using recursion
import java.util.ArrayList;
import java.util.List;
class GfG {
static void findSubsets(int[] arr, int index,
int target, List<Integer> curr,
List<List<Integer>> result) {
if (index >= arr.length) {
// If we reach the end and the target
// becomes 0, we found a valid subset
if (target == 0) {
result.add(new ArrayList<>(curr));
return;
}
// Otherwise, we return as no valid
// subset is found
return;
}
// Include current element in subset
curr.add(arr[index]);
findSubsets(arr, index + 1,
target - arr[index], curr, result);
// Backtrack and exclude the current element
curr.remove(curr.size() - 1);
findSubsets(arr,
index + 1, target, curr, result);
}
// Function to find all subsets summing to target
static List<List<Integer>> perfectSum(int[] arr,
int target) {
List<List<Integer>> result = new ArrayList<>();
List<Integer> curr = new ArrayList<>();
findSubsets(arr, 0, target, curr, result);
return result;
}
// Function to print subsets in required format
static void print2dArray(List<List<Integer>> arr) {
if (arr.isEmpty()) {
// No valid subsets found
System.out.println("-1");
return;
}
for (int row = 0; row < arr.size(); row++) {
System.out.print("[");
for (int col = 0; col < arr.get(row).size(); col++) {
System.out.print(arr.get(row).get(col));
if (col != arr.get(row).size() - 1) {
System.out.print(", ");
}
}
System.out.print("]");
if (row < arr.size() - 1) System.out.print(", ");
}
}
public static void main(String[] args) {
int[] arr = {5, 2, 3, 10, 6, 8};
int target = 10;
// Find subsets and print result
List<List<Integer>> result = perfectSum(arr, target);
print2dArray(result);
}
}
# Python Code to find subsets with sum equal
# to target using recursion
def findSubsets(arr, index, target, curr, result):
if index >= len(arr):
# If we reach the end and the target
# becomes 0, we found a valid subset
if target == 0:
result.append(curr[:])
return
# Otherwise, we return as no valid
# subset is found
return
# Include current element in subset
curr.append(arr[index])
findSubsets(arr, index + 1, target - arr[index], curr, result)
# Backtrack and exclude the current element
curr.pop()
findSubsets(arr, index + 1, target, curr, result)
# Function to find all subsets summing to target
def perfectSum(arr, target):
result = []
curr = []
findSubsets(arr, 0, target, curr, result)
return result
# Function to print subsets in required format
def print2dArray(arr):
if not arr:
# No valid subsets found
print("-1")
return
for row in range(len(arr)):
print("[", end="")
for col in range(len(arr[row])):
print(arr[row][col], end="")
if col != len(arr[row]) - 1:
print(", ", end="")
print("]", end="")
if row < len(arr) - 1:
print(", ", end="")
if __name__ == "__main__":
arr = [5, 2, 3, 10, 6, 8]
target = 10
# Find subsets and print result
result = perfectSum(arr, target)
print2dArray(result)
// C# Code to find subsets with sum equal
// to target using recursion
using System;
using System.Collections.Generic;
class GfG {
static void findSubsets(int[] arr, int index,
int target, List<int> curr,
List<List<int>> result) {
if (index >= arr.Length) {
// If we reach the end and the target
// becomes 0, we found a valid subset
if (target == 0) {
result.Add(new List<int>(curr));
return;
}
// Otherwise, we return as no valid
// subset is found
return;
}
// Include current element in subset
curr.Add(arr[index]);
findSubsets(arr, index + 1,
target - arr[index], curr, result);
// Backtrack and exclude the current element
curr.RemoveAt(curr.Count - 1);
findSubsets(arr,
index + 1, target, curr, result);
}
// Function to find all subsets summing to target
static List<List<int>> perfectSum(int[] arr,
int target) {
List<List<int>> result = new List<List<int>>();
List<int> curr = new List<int>();
findSubsets(arr, 0, target, curr, result);
return result;
}
// Function to print subsets in required format
static void print2dArray(List<List<int>> arr) {
if (arr.Count == 0) {
// No valid subsets found
Console.WriteLine("-1");
return;
}
for (int row = 0; row < arr.Count; row++) {
Console.Write("[");
for (int col = 0; col < arr[row].Count; col++) {
Console.Write(arr[row][col]);
if (col != arr[row].Count - 1) {
Console.Write(", ");
}
}
Console.Write("]");
if (row < arr.Count - 1) Console.Write(", ");
}
}
static void Main() {
int[] arr = {5, 2, 3, 10, 6, 8};
int target = 10;
// Find subsets and print result
List<List<int>> result = perfectSum(arr, target);
print2dArray(result);
}
}
// JavaScript Code to find subsets with sum equal
// to target using recursion
function findSubsets(arr, index, target, curr, result) {
if (index >= arr.length) {
// If we reach the end and the target
// becomes 0, we found a valid subset
if (target === 0) {
result.push([...curr]);
return;
}
// Otherwise, we return as no valid
// subset is found
return;
}
// Include current element in subset
curr.push(arr[index]);
findSubsets(arr, index + 1, target - arr[index], curr, result);
// Backtrack and exclude the current element
curr.pop();
findSubsets(arr, index + 1, target, curr, result);
}
// Function to find all subsets summing to target
function perfectSum(arr, target) {
let result = [];
let curr = [];
findSubsets(arr, 0, target, curr, result);
return result;
}
// Function to print subsets in required format
function print2dArray(arr) {
if (arr.length === 0) {
// No valid subsets found
console.log("-1");
return;
}
let output = arr.map(subset => "[" + subset.join(", ") + "]").join(", ");
console.log(output);
}
let arr = [5, 2, 3, 10, 6, 8];
let target = 10;
// Find subsets and print result
let result = perfectSum(arr, target);
print2dArray(result);
Output[5, 2, 3], [2, 8], [10]
Time Complexity: O(2^n), as each element has two choices, include or exclude.
Space Complexity: O(n), due to maximum recursion depth in the worst case scenario.
Using Dynamic Programming – O(2^n) Time and O(n*target) Space
The idea is to use Dynamic Programming (DP) to determine whether a subset with the given target sum exists. We build a dp table where dp[i][j] indicates if a sum j is possible using the first i elements. Once the table is constructed, we use recursion to backtrack and find all valid subsets. If 0s are present, we handle them separately to ensure all possible combinations are included.
Steps to implement the above idea:
- Initialize a dp table of size n × (target + 1) to track subset sum possibilities using boolean values.
- Set dp[i][0] = true for all i since a sum of 0 is always possible with an empty subset.
- Fill the dp table by checking whether a sum j can be formed by including or excluding arr[i].
- If dp[n-1][target] is false, return an empty result as no valid subset exists.
- Use recursion to backtrack through the dp table and find all subsets summing to target.
- Maintain a temporary list to store the current subset and push valid ones to the result list.
- Finally, print the result list in the required format, handling cases where no valid subset exists.
Below is an implementation of the above approach:
C++
// C++ Code to find subsets with sum equal to target
// using Dynamic Programming
#include <bits/stdc++.h>
using namespace std;
// Recursively finds all subsets with the given target
void findSubsets(vector<int>& arr, vector<vector<bool>>& dp,
int i, int target, vector<int>& curr,
vector<vector<int>>& res) {
// Base case: If target becomes 0
if (i == 0) {
if (target == 0) res.push_back(curr);
if (arr[0] == target) {
curr.push_back(arr[0]);
res.push_back(curr);
curr.pop_back();
}
return;
}
// Exclude current element
if (dp[i-1][target]) {
findSubsets(arr, dp, i-1, target, curr, res);
}
// Include current element if it does not exceed target
if (target >= arr[i] && dp[i-1][target-arr[i]]) {
curr.push_back(arr[i]);
findSubsets(arr, dp, i-1, target-arr[i], curr, res);
curr.pop_back();
}
}
// Returns all subsets with the given target sum
vector<vector<int>> perfectSum(vector<int>& arr,
int target) {
int n = arr.size();
if (n == 0 || target < 0) return {};
// DP table to store subset sum possibilities
vector<vector<bool>> dp(n, vector<bool>(target+1, false));
// Correct DP initialization for handling zeroes
dp[0][0] = true;
if (arr[0] <= target) dp[0][arr[0]] = true;
for (int i = 1; i < n; ++i) {
for (int j = 0; j <= target; ++j) {
dp[i][j] = dp[i-1][j]
|| (arr[i] <= j && dp[i-1][j-arr[i]]);
}
}
// If no subsets sum to target, return empty
if (!dp[n-1][target]) return {};
vector<vector<int>> res;
vector<int> curr;
findSubsets(arr, dp, n-1, target, curr, res);
return res;
}
// Function to print subsets in required format
void print2dArray(vector<vector<int>>& arr) {
if (arr.empty()) {
// No valid subsets found
cout << "-1\n";
return;
}
// Printing subsets in formatted output
for (int row = 0; row < arr.size(); row++) {
cout << "[";
for (int col = arr[row].size() - 1; col >= 0 ; col--) {
cout << arr[row][col];
if (col != 0) {
cout << ", ";
}
}
cout << "]";
if (row < arr.size() - 1) cout << ", ";
}
}
// Driver function
int main() {
vector<int> arr = {5, 2, 3, 10, 6, 8};
int target = 10;
vector<vector<int>> result = perfectSum(arr, target);
print2dArray(result);
return 0;
}
// Java Code to find subsets with sum equal to target
// using Dynamic Programming
import java.util.*;
class GfG {
// Recursively finds all subsets with the given target
static void findSubsets(int[] arr, boolean[][] dp,
int i, int target, List<Integer> curr,
List<List<Integer>> res) {
// Base case: If target becomes 0
if (i == 0) {
if (target == 0) res.add(new ArrayList<>(curr));
if (arr[0] == target) {
curr.add(arr[0]);
res.add(new ArrayList<>(curr));
curr.remove(curr.size() - 1);
}
return;
}
// Exclude current element
if (dp[i-1][target]) {
findSubsets(arr, dp, i-1, target, curr, res);
}
// Include current element if it does not exceed target
if (target >= arr[i] && dp[i-1][target-arr[i]]) {
curr.add(arr[i]);
findSubsets(arr, dp, i-1, target-arr[i], curr, res);
curr.remove(curr.size() - 1);
}
}
// Returns all subsets with the given target sum
static List<List<Integer>> perfectSum(int[] arr, int target) {
int n = arr.length;
if (n == 0 || target < 0) return new ArrayList<>();
// DP table to store subset sum possibilities
boolean[][] dp = new boolean[n][target + 1];
// Correct DP initialization for handling zeroes
dp[0][0] = true;
if (arr[0] <= target) dp[0][arr[0]] = true;
for (int i = 1; i < n; ++i) {
for (int j = 0; j <= target; ++j) {
dp[i][j] = dp[i-1][j]
|| (arr[i] <= j && dp[i-1][j-arr[i]]);
}
}
// If no subsets sum to target, return empty
if (!dp[n-1][target]) return new ArrayList<>();
List<List<Integer>> res = new ArrayList<>();
findSubsets(arr, dp, n-1, target, new ArrayList<>(), res);
return res;
}
// Function to print subsets in required format
static void print2dArray(List<List<Integer>> arr) {
if (arr.isEmpty()) {
// No valid subsets found
System.out.println("-1");
return;
}
// Printing subsets in formatted output
for (int row = 0; row < arr.size(); row++) {
System.out.print("[");
for (int col = arr.get(row).size() - 1; col >= 0; col--) {
System.out.print(arr.get(row).get(col));
if (col != 0) {
System.out.print(", ");
}
}
System.out.print("]");
if (row < arr.size() - 1) System.out.print(", ");
}
}
// Driver function
public static void main(String[] args) {
int[] arr = {5, 2, 3, 10, 6, 8};
int target = 10;
List<List<Integer>> result = perfectSum(arr, target);
print2dArray(result);
}
}
# Python Code to find subsets with sum equal to target
# using Dynamic Programming
# Recursively finds all subsets with the given target
def findSubsets(arr, dp, i, target, curr, res):
# Base case: If target becomes 0
if i == 0:
if target == 0:
res.append(curr[:])
if arr[0] == target:
curr.append(arr[0])
res.append(curr[:])
curr.pop()
return
# Exclude current element
if dp[i-1][target]:
findSubsets(arr, dp, i-1, target, curr, res)
# Include current element if it does not exceed target
if target >= arr[i] and dp[i-1][target-arr[i]]:
curr.append(arr[i])
findSubsets(arr, dp, i-1, target-arr[i], curr, res)
curr.pop()
# Returns all subsets with the given target sum
def perfectSum(arr, target):
n = len(arr)
if n == 0 or target < 0:
return []
# DP table to store subset sum possibilities
dp = [[False] * (target+1) for _ in range(n)]
# Correct DP initialization for handling zeroes
dp[0][0] = True
if arr[0] <= target:
dp[0][arr[0]] = True
for i in range(1, n):
for j in range(target+1):
dp[i][j] = dp[i-1][j] or (arr[i] <= j and dp[i-1][j-arr[i]])
# If no subsets sum to target, return empty
if not dp[n-1][target]:
return []
res = []
curr = []
findSubsets(arr, dp, n-1, target, curr, res)
return res
# Function to print subsets in required format
def print2dArray(arr):
if not arr:
# No valid subsets found
print("-1")
return
# Printing subsets in formatted output
for row in range(len(arr)):
print("[", end="")
for col in range(len(arr[row]) - 1, -1, -1):
print(arr[row][col], end="")
if col != 0:
print(", ", end="")
print("]", end="")
if row < len(arr) - 1:
print(", ", end="")
if __name__ == "__main__":
arr = [5, 2, 3, 10, 6, 8]
target = 10
result = perfectSum(arr, target)
print2dArray(result)
// C# Code to find subsets with sum equal to target
// using Dynamic Programming
using System;
using System.Collections.Generic;
class GfG {
// Recursively finds all subsets with the given target
static void findSubsets(int[] arr, bool[,] dp,
int i, int target, List<int> curr,
List<List<int>> res) {
// Base case: If target becomes 0
if (i == 0) {
if (target == 0) res.Add(new List<int>(curr));
if (arr[0] == target) {
curr.Add(arr[0]);
res.Add(new List<int>(curr));
curr.RemoveAt(curr.Count - 1);
}
return;
}
// Exclude current element
if (dp[i-1, target]) {
findSubsets(arr, dp, i-1, target, curr, res);
}
// Include current element if it does not exceed target
if (target >= arr[i] && dp[i-1, target-arr[i]]) {
curr.Add(arr[i]);
findSubsets(arr, dp, i-1, target-arr[i], curr, res);
curr.RemoveAt(curr.Count - 1);
}
}
// Returns all subsets with the given target sum
static List<List<int>> perfectSum(int[] arr, int target) {
int n = arr.Length;
if (n == 0 || target < 0) return new List<List<int>>();
// DP table to store subset sum possibilities
bool[,] dp = new bool[n, target + 1];
// Correct DP initialization for handling zeroes
dp[0, 0] = true;
if (arr[0] <= target) dp[0, arr[0]] = true;
for (int i = 1; i < n; ++i) {
for (int j = 0; j <= target; ++j) {
dp[i, j] = dp[i-1, j]
|| (arr[i] <= j && dp[i-1, j-arr[i]]);
}
}
// If no subsets sum to target, return empty
if (!dp[n-1, target]) return new List<List<int>>();
List<List<int>> res = new List<List<int>>();
findSubsets(arr, dp, n-1, target, new List<int>(), res);
return res;
}
// Function to print subsets in required format
static void print2dArray(List<List<int>> arr) {
if (arr.Count == 0) {
// No valid subsets found
Console.WriteLine("-1");
return;
}
// Printing subsets in formatted output
for (int row = 0; row < arr.Count; row++) {
Console.Write("[");
for (int col = arr[row].Count - 1; col >= 0; col--) {
Console.Write(arr[row][col]);
if (col != 0) {
Console.Write(", ");
}
}
Console.Write("]");
if (row < arr.Count - 1) Console.Write(", ");
}
}
// Driver function
static void Main() {
int[] arr = {5, 2, 3, 10, 6, 8};
int target = 10;
List<List<int>> result = perfectSum(arr, target);
print2dArray(result);
}
}
// JavaScript Code to find subsets with sum equal to target
// using Dynamic Programming
function findSubsets(arr, dp, i, target, curr, res) {
// Base case: If target becomes 0
if (i === 0) {
if (target === 0) res.push([...curr]);
if (arr[0] === target) {
curr.push(arr[0]);
res.push([...curr]);
curr.pop();
}
return;
}
// Exclude current element
if (dp[i - 1][target]) {
findSubsets(arr, dp, i - 1, target, curr, res);
}
// Include current element if it does not exceed target
if (target >= arr[i] && dp[i - 1][target - arr[i]]) {
curr.push(arr[i]);
findSubsets(arr, dp, i - 1, target - arr[i], curr, res);
curr.pop();
}
}
function perfectSum(arr, target) {
let n = arr.length;
if (n === 0 || target < 0) return [];
// DP table to store subset sum possibilities
let dp = Array.from({ length: n }, () => Array(target + 1).fill(false));
// Correct DP initialization for handling zeroes
dp[0][0] = true;
if (arr[0] <= target) dp[0][arr[0]] = true;
for (let i = 1; i < n; ++i) {
for (let j = 0; j <= target; ++j) {
dp[i][j] = dp[i - 1][j] || (arr[i] <= j && dp[i - 1][j - arr[i]]);
}
}
// If no subsets sum to target, return empty
if (!dp[n - 1][target]) return [];
let res = [];
let curr = [];
findSubsets(arr, dp, n - 1, target, curr, res);
return res;
}
function print2dArray(arr) {
if (arr.length === 0) {
// No valid subsets found
console.log("-1");
return;
}
// Printing subsets in formatted output
let output = "";
for (let row = 0; row < arr.length; row++) {
output += "[";
for (let col = arr[row].length - 1; col >= 0; col--) {
output += arr[row][col];
if (col !== 0) {
output += ", ";
}
}
output += "]";
if (row < arr.length - 1) output += ", ";
}
console.log(output);
}
// Driver Code
let arr = [5, 2, 3, 10, 6, 8];
let target = 10;
let result = perfectSum(arr, target);
print2dArray(result);
Output[5, 2, 3], [10], [2, 8]
Time Complexity: O(2^n), as DP table filling takes O(n * target), and subset generation takes O(2^n).
Space Complexity: O(n * target), as DP table uses O(n * target) space, and recursion stack takes O(n).
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Subset Sum Problem using Backtracking
Given a set[] of non-negative integers and a value sum, the task is to print the subset of the given set whose sum is equal to the given sum. Examples:Â Input:Â set[] = {1,2,1}, sum = 3Output:Â [1,2],[2,1]Explanation:Â There are subsets [1,2],[2,1] with sum 3. Input:Â set[] = {3, 34, 4, 12, 5, 2}, sum =
8 min read
Print all subsets with given sum
Given an array arr[] of non-negative integers and an integer target. The task is to print all subsets of the array whose sum is equal to the given target. Note: If no subset has a sum equal to target, print -1. Examples: Input: arr[] = [5, 2, 3, 10, 6, 8], target = 10Output: [ [5, 2, 3], [2, 8], [10
15+ min read
Subset Sum Problem in O(sum) space
Given an array of non-negative integers and a value sum, determine if there is a subset of the given set with sum equal to given sum. Examples: Input: arr[] = {4, 1, 10, 12, 5, 2}, sum = 9Output: TRUEExplanation: {4, 5} is a subset with sum 9. Input: arr[] = {1, 8, 2, 5}, sum = 4Output: FALSE Explan
13 min read
Subset Sum is NP Complete
Prerequisite: NP-Completeness, Subset Sum Problem Subset Sum Problem: Given N non-negative integers a1...aN and a target sum K, the task is to decide if there is a subset having a sum equal to K. Explanation: An instance of the problem is an input specified to the problem. An instance of the subset
5 min read
Minimum Subset sum difference problem with Subset partitioning
Given a set of N integers with up to 40 elements, the task is to partition the set into two subsets of equal size (or the closest possible), such that the difference between the sums of the subsets is minimized. If the size of the set is odd, one subset will have one more element than the other. If
13 min read
Maximum subset sum such that no two elements in set have same digit in them
Given an array of N elements. Find the subset of elements which has maximum sum such that no two elements in the subset has common digit present in them.Examples: Input : array[] = {22, 132, 4, 45, 12, 223} Output : 268 Maximum Sum Subset will be = {45, 223} . All possible digits are present except
12 min read
Find all distinct subset (or subsequence) sums of an array
Given an array arr[] of size n, the task is to find a distinct sum that can be generated from the subsets of the given sets and return them in increasing order. It is given that the sum of array elements is small. Examples: Input: arr[] = [1, 2]Output: [0, 1, 2, 3]Explanation: Four distinct sums can
15+ min read
Subset sum problem where Array sum is at most N
Given an array arr[] of size N such that the sum of all the array elements does not exceed N, and array queries[] containing Q queries. For each query, the task is to find if there is a subset of the array whose sum is the same as queries[i]. Examples: Input: arr[] = {1, 0, 0, 0, 0, 2, 3}, queries[]
10 min read